Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int N = in.nextInt();
		double[] dp = new double[N+1];
		dp[0] = 1;
		double ans = 0.0;
		
		for(int coins = 0; coins < N; coins++) {
		    double prob = in.nextDouble();
		    for(int j = coins+1; j >= 0; j--) {
		        dp[j] = (j == 0 ? 0 : dp[j-1]*prob) + dp[j]*(1-prob);
		    }
		}
		
		for(int heads = 0; heads <= N; heads++) {
		    int tails = N - heads;
		    if(heads > tails) {
		        ans += dp[heads];
		    }
		}
		
		System.out.println(ans);
	}
}